Wave propagation in elastic tubes.

[somasimple]

Hi All,

Here is my question regarding the solution of this little movie =>
http://www.somasimple.com/flash_anims/ap_003.swf

A thin elastic tube (thickness = 1) is connected with a thicker one (40 times thicker than the previous section). The tubes are poured with water.
A wave is is traveling in the first section changing the shape of the tube (I emphasized the deformation on the movie). Its speed (constant) is known and equals to c1.

What happens when the wave reaches the thicker tube?
It is supposed that thre is no wave reflection.

Is c2 > c1?
How is the amplitude changed?

 

[FredGarvin]

Step discontinuities like this are a bit different than what I am used to, but I would imagine that this would get you in the ballpark of where you need to be. However, looking at what you have provided, I'm not convinced you have enough information provided to answer the question.

At the discontinuity, there will be a reflected and a transmitted wave due to the incident wave. You can get a feel for the relative values by looking at the mechanical impedance of each tube.

Set the small tube to be T1 and the larger to be T2. This would result in Z1 and Z2 as the two corresponding impedances, where

$$Z = \rho c$$ where
$$\rho$$=density
$$c$$=speed of sound in the medium

The incident wave is usually described in terms of
$$u_i(x,t)=A_ie^{i(\omega t-kx)}$$

The reflected and transmitted waves are in the form of
$$u_r(x,t)=A_re^{i(\omega t+kx)}$$
$$u_t(x,t)=A_te^{i(\omega t-kx)}$$

So, the expression that relates the A coefficients looks like:

$$\frac{A_t}{A_i}=\frac{2Z_1}{Z_1+Z_2}$$

This leads to the fact that you must either already know the speed of sound in both media or you must know something about either the transmitted or reflected waves.

 

[somasimple]

Hi,

Thanks for the reply!
It is is just a virtual hypothesis.

With the enounced problem may I conclude that amplitude is lowered because impedance Z2>Z1?